Question: In the diagram below, we have $AB = 24$ and $\angle ADB =90^\circ$.  If $\sin A = \frac23$ and $\sin C = \frac13$, then what is $DC$?


[asy]
pair A,B,C,D;
A = (0,0);
B = (8*sqrt(5),16);
D = (8*sqrt(5),0);
C = (8*sqrt(5) + 32*sqrt(2),0);
draw(D--B--A--C--B);
label("$A$",A,SW);
label("$B$",B,N);
label("$C$",C,SE);
label("$D$",D,S);
draw(rightanglemark(B,D,A,63));
[/asy]
Solution: From right triangle $ABD$, we have $\sin A = \frac{BD}{AB} = \frac{BD}{24}$.  Since $\sin A = \frac23$, we have $\frac23 = \frac{BD}{24}$, so $BD = \frac23\cdot 24 = 16$.

From right triangle $BCD$, we have $\sin C = \frac{BD}{BC}=\frac{16}{BC}$.  Since $\sin C = \frac13$, we have $\frac{16}{BC} = \frac13$.  Therefore, we have $BC = 3\cdot 16=48$.  Finally, the Pythagorean Theorem gives us \begin{align*}
CD &= \sqrt{BC^2 - BD^2}\\
&= \sqrt{48^2 - 16^2} \\
&= \sqrt{(3\cdot 16)^2 - 16^2} \\
&= \sqrt{9\cdot 16^2 - 16^2} = \sqrt{8\cdot 16^2} = 2\cdot 16 \sqrt{2} = \boxed{32\sqrt{2}}.\end{align*}